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whenever x,y ∈ X, 0 ≤ s ≤ t ≤ 1, p := σxy(s), q := σxy(t), and λ ∈ [0, 1]. Thus, every conical and consistent geodesic bicombing is convex. Furthermore, we will say that σ is reversible if

σxy(t)= σyx(1 − t) for all t ∈ [0, 1] (1.4) and x,y ∈ X. This property is not imposed here but will be obtained at no extra cost later. Basic examples of convex geodesic bicombings are given by the linear geodesics σxy(t) := (1 − t)x + ty in a linearly convex subset X of a normed space or by the unique geodesics σxy : [0, 1] → X in a CAT(0) space [2, 4] or a Busemann space [1, 20] (where t 7→ d(σ(t),τ(t)) is convex for every pair of geodesics σ, τ : [0, 1] → X.) An additional source of conical geodesic bicombings is the fact that this weaker notion behaves well under 1-Lipschitz retractions (see Lemma 2.1). Consistent and reversible geodesic bicombings have also been employed in [13, 10]. In view of the primary examples, the existence of a convex or conical geodesic bicombing on a may be regarded as a weak (but non-coarse) global nonpositive curvature condition. One thus arrives at the following hierarchy of properties (A) ⇒ (B) ⇒ (C) ⇒ (D) ⇒ (E) for a geodesic metric space X:

(A) X is a CAT(0) space;

(B) X is a Busemann space;

(C) X admits a convex and consistent geodesic bicombing;

(D) X admits a convex geodesic bicombing;

(E) X admits a conical geodesic bicombing.

Clearly, if X is uniquely geodesic, then (E) ⇒ (B). When X is a normed real vector space, (A) holds if and only if the norm is induced by an inner product ([4, Proposition II.1.14]), (B) holds if and only if the norm is strictly convex ([20, Proposition 8.1.6]), and (C) is always satisfied. In the general case, (C) is stable under limit operations, whereas (B) is not (compare Sect. 10 in [17]). It is unclear whether (E) ⇒ (D) and (D) ⇒ (C) without further conditions. One of the purposes of this paper is to establish these implications under suitable assumptions, and to address questions of uniqueness. Our first result refers to (E) and (D).

1.1 Theorem. Let X be a proper metric space with a conical geodesic bicombing. Then X also admits a convex geodesic bicombing.

The idea of the proof is to resort to a discretized convexity condition and then to gradually decrease the parameter of discreteness by the “cat’s cradle” construction from [1]. The passage from (D) to (C) appears to be more subtle. We first observe that the linear segments t 7→ (1 − t)x + ty in an arbitrary normed Convex geodesic bicombings and hyperbolicity 3 space X may be characterized as the curves γ : [0, 1] → X with the property that t 7→ d(z, γ(t)) is convex on [0, 1] for every z ∈ X (see Theorem 3.3). We therefore term curves with this property straight (in a metric space X). Since the geodesics of a convex bicombing are necessarily straight, the above observation shows that normed spaces have no such bicombing other than the linear one. By contrast, there are instances of non-linear straight geodesics in compact convex subspaces of normed spaces as well as multiple convex geodesic bicombings in compact metric spaces (see Examples 3.4 and 3.5). Nevertheless, we prove a strong uniqueness property of straight curves (Proposition 4.3) in spaces satisfying a certain finite dimensionality assumption, introduced by Dress in [7] and explained further below. This gives the following result regarding items (D) and (C).

1.2 Theorem. Let X be a metric space of finite combinatorial dimension in the sense of Dress, or with the property that every bounded subset has finite com- binatorial dimension. Suppose that X possesses a convex geodesic bicombing σ. Then σ is consistent, reversible, and unique, that is, σ is the only convex geodesic bicombing on X.

Our interest in Theorems 1.1 and 1.2 comes from the fact that property (E) holds in particular for all absolute 1-Lipschitz retracts X. (To see this, embed X isometrically into ℓ∞(X) and retract the linear geodesics to X; compare again Lemma 2.1). These correspond to the injective objects in the category of metric spaces and 1-Lipschitz maps: X is injective if for every isometric inclusion A ⊂ B of metric spaces and every 1-Lipschitz map f : A → X there exists a 1-Lipschitz extension f¯: B → X of f. Basic examples include the real line, all ℓ∞ spaces, and all metric (R-)trees. Furthermore, by a 50-year-old result of Isbell [14], every metric space X possesses an essentially unique injective hull E(X). This yields a compact metric space if X is compact and a finite polyhedral complex with ℓ∞ metrics on the cells in case X is finite. Isbell’s explicit construction was redis- covered and further explored by Dress [7], who gave it the name tight span. The combinatorial dimension dimcomb(X) of a general metric space X is the supremum of the dimensions of the polyhedral complexes E(Y ) for all finite subspaces Y of X. In case X is already injective, every such E(Y ) embeds isometrically into X, so dimcomb(X) is bounded by the supremum of the topological dimensions of com- pact subsets of X. Dress also gave a characterization of the condition dimcomb ≤ n in terms of a 2(n + 1)-point inequality. We restate his result in Theorem 4.1 and provide a streamlined proof in an appendix to this paper. In [18], the second author proved that if Γ is a Gromov hyperbolic group, endowed with the word metric with respect to some finite generating set, then the injective hull E(Γ) is a proper, finite-dimensional polyhedral complex with finitely many types of cells, isometric to polytopes in ℓ∞ spaces, and Γ acts properly and cocompactly on E(Γ) by cellular . Furthermore, after barycentric subdivision, the resulting simplicial Γ-complex is a model for the classifying space EΓ for proper actions. Since X = E(Γ) satisfies property (E), 4 D. Descombes & U. Lang

Theorems 1.1 and 1.2 together now show that E(Γ) possesses a convex geodesic bicombing that is consistent, reversible, and unique, hence equivariant with re- spect to the full isometry group of E(Γ). (The existence of an equivariant conical geodesic bicombing on E(Γ) was known before, see Proposition 3.8 in [18], but the proof of that fact did not give any indication on the consistency and uniqueness of the bicombing.) In particular, the following holds. 1.3 Theorem. Every word hyperbolic group Γ acts properly and cocompactly by isometries on a proper, finite-dimensional metric space X with a convex geodesic bicombing that is furthermore consistent, reversible, and unique, hence equivariant with respect to the isometry group of X. In a last part of the paper, we discuss the asymptotic geometry of complete metric spaces X with a convex and consistent geodesic bicombing σ. We define the geometric boundary ∂σX in terms of geodesic rays consistent with σ, and we equip Xσ = X ∪∂σX with a natural metrizable topology akin to the cone topology in the case of CAT(0) spaces. In view of Theorem 1.3, this unifies and generalizes the respective constructions for CAT(0) or Busemann spaces and for hyperbolic groups. By virtue of the bicombing one can then give a rather direct proof of the following result on Xσ. Recall that a metrizable space is an absolute retract if it is a retract of every metrizable space containing it as a closed subspace. 1.4 Theorem. Let X be a with a convex and consistent geodesic bicombing σ. Then Xσ is contractible, locally contractible, and an abso- lute retract. Moreover, ∂σX is a Z-set in Xσ, that is, for every open set U in Xσ .the inclusion U \ ∂σX ֒→ U is a homotopy equivalence Note that we do not assume X to be proper or finite dimensional. Bestvina and Mess [3] proved the analogous result for the Gromov closure P (Γ) of the (contractible) Rips complex of a hyperbolic group Γ. Theorem 1.3 and Theo- rem 1.4 together thus provide an alternative to their result, which has a number of important applications (see Corollaries 1.3 and 1.4 in [3]).

2 From conical to convex bicombings

In this section, we first discuss some simple properties and examples of conical geodesic bicombings, as defined in (1.2), then we prove Theorem 1.1. 2.1 Lemma. Let X¯ be a metric space with a conical geodesic bicombing σ¯. If π : X¯ → X is a 1-Lipschitz retraction onto some subspace X of X¯, then σ := π ◦ σ¯|X×X×[0,1] defines a conical geodesic bicombing on X. Proof. Note that since π is a 1-Lipschitz retraction, σ is indeed a geodesic bicomb- ing on X. Furthermore, since π is 1-Lipschitz andσ ¯ is conical, we have ′ ′ d(σxy(t),σx′y′ (t)) ≤ d(¯σxy(t), σ¯x′y′ (t)) ≤ (1 − t) d(x,x )+ t d(y,y ) for all t ∈ [0, 1] and x,y,x′,y′ ∈ X, so σ is conical as well. Convex geodesic bicombings and hyperbolicity 5

As mentioned earlier, a direct consequence of this lemma is that all injective metric spaces (or absolute 1-Lipschitz retracts) admit conical geodesic bicombings. An equally simple application gives an example of a conical geodesic bicombing that is not convex.

2.2 Example. Consider the set X of all (u, v) ∈ R2 with |u| ≤ 2 and b(u) := 2 |u|− 1 ≤ v ≤ |b(u)|, endowed with the metric induced by the ℓ∞ norm on R . Note that this is a geodesic metric space. With respect to this metric, the vertical retraction π from the triangle X¯ := {(u, v) : b(u) ≤ v ≤ 1} onto X that maps (u, v) to (u, min{v, |b(u)|}) is 1-Lipschitz. The linear geodesic bicombingσ ¯ on X¯, defined byσ ¯xy(t) := (1 − t)x + ty, is convex. By Lemma 2.1, σ := π ◦ σ¯|X×X×[0,1] defines a conical geodesic bicombing on X. For x = (−2, 1) and y = (2, 1), we have σxy(1/4) = (−1, 0), σxy(3/4) = (1, 0), and σxy(1/2) = (0, 1). Hence, for z = (0, −1), the function t 7→ kσxy(t) − σzz(t)k∞ = kσxy(t) − zk∞ is clearly not convex.

We now define a relaxed notion of convexity that will be useful for the proof of Theorem 1.1. We say that a geodesic bicombing σ on a metric space X is 1/n-discretely convex if, for all x,y,x′,y′ ∈ X, the convexity condition t − s s − r d(σ (s),σ ′ ′ (s)) ≤ d(σ (r),σ ′ ′ (r)) + d(σ (t),σ ′ ′ (t)) xy x y t − r xy x y t − r xy x y holds whenever the three numbers r

2d(σxy(s),σx′y′ (s)) ≤ d(σxy(s − 1/n),σx′y′ (s − 1/n))

+ d(σxy(s + 1/n),σx′y′ (s + 1/n)) for all s ∈ (0, 1) ∩ (1/n)Z. Note that every conical geodesic bicombing is 1/2- discretely convex.

2.3 Proposition. Suppose that X is a complete metric space with a geodesic bicombing σ that is conical and 1/n-discretely convex for some integer n ≥ 2. Then X also admits a geodesic bicombing that is conical and 1/(2n − 1)-discretely convex.

Proof. Set m := 2n − 1. To construct the desired new bicombingσ ˜, fix x and y and define the sequences pi and qi recursively as q0 := σxy(n/m) and

pi := σxqi−1 (1 − 1/n), qi := σpiy(1/n), for i ≥ 1.

Since σ is conical, we get the inequalities

d(pi,pi+1) ≤ (1 − 1/n) d(qi−1,qi),

d(qi,qi+1) ≤ (1 − 1/n) d(pi,pi+1). 6 D. Descombes & U. Lang

It follows that pi and qi are Cauchy sequences, so pi → p and qi → q. Then

σxqi → σxq and σpiy → σpy uniformly, again because σ is conical. Note that p = σxq(1−1/n) and q = σpy(1/n); we can thus defineσ ˜xy(s) for s ∈ [0, 1]∩(1/m)Z so that σxq (m/n)s if s ≤ n/m, σ˜xy(s)= (2.1) (σpy(m/n)s− (n − 1)/n if s ≥ (n − 1)/m. ′ To declareσ ˜xy on all of [0, 1], we connect any pair of consecutive points x :=σ ˜xy(s) ′ and y :=σ ˜xy(s+1/m), where s ∈ [0, 1)∩(1/m)Z, by the geodesic t 7→ σx′y′ (m(t− s)) for t ∈ [s,s + 1/m]. ′ ′ ′ Now if p ,q andσ ˜x′y′ result from the same construction for two points x and y′, we want to show that for s ∈ (0, 1) ∩ (1/m)Z we have

2d(˜σxy(s), σ˜x′y′ (s)) ≤ d(˜σxy(s − 1/m), σ˜x′y′ (s − 1/m))

+ d(˜σxy(s + 1/m), σ˜x′y′ (s + 1/m)).

In view of (2.1) this corresponds to the inequality

2d(τ(t),τ ′(t)) ≤ d(τ(t − 1/n),τ ′(t − 1/n)) + d(τ(t + 1/n),τ ′(t + 1/n))

′ ′ where τ = σxq, τ = σx′q′ , t = (m/n)s if s ≤ (n − 1)/m and τ = σpy, τ = σp′y′ , t = (m/n)s − (n − 1)/n if s ≥ n/m. However, these inequalities hold since σ is 1/n-discretely convex. This shows thatσ ˜ is 1/m-discretely convex. Now it follows easily from the construction thatσ ˜ is also conical.

From this result, we now obtain Theorem 1.1 by an application of the Arzel`a– Ascoli theorem, which requires X to be proper. We do not know whether the implication (E) ⇒ (D) holds in general without this assumption.

Proof of Theorem 1.1. Starting from the given conical geodesic bicombing σ1 := σ, we construct, by means of Proposition 2.3, a sequence of conical geodesic bi- k k combings σ on X such that σ is 1/nk-discretely convex, where n1 = 2 and k nk+1 = 2nk − 1. This collection of maps σ is equicontinuous on every bounded set, and for every fixed (x,y,t) in the separable domain X ×X ×[0, 1], the sequence k σxy(t) remains in a compact subset of X. One may thus extract a subsequence σk(l) that converges uniformly on every compact set to a mapσ ¯, which is clearly a geodesic bicombing. Convexity t − s s − r d(¯σ (s), σ¯ ′ ′ (s)) ≤ d(¯σ (r), σ¯ ′ ′ (r)) + d(¯σ (t), σ¯ ′ ′ (t)), xy x y t − r xy x y t − r xy x y

k(l) k(l) where r

2.4 Remark. Let σ be a concial geodesic bicombing on the proper metric space X, and suppose that for some pair of points x,y the consistency condition σx′y′ (λ)= ′ σxy((1 − λ)s + λt) holds for all 0 ≤ s ≤ t ≤ 1 and λ ∈ [0, 1], where x := σxy(s) ′ and y := σxy(t). Then it is easily seen that σxy as well as all positively oriented subsegments are unaltered by the procedure in the above proof. In other words, ′ ′ the resulting convex bicombingσ ¯ satisfiesσ ¯x′y′ = σx′y′ for all x ,y as above, in particularσ ¯xy = σxy.

3 Straight curves

Whereas the preceding section dealt with the existence of convex bicombings, we now turn to the question of uniqueness. First we consider a property every geodesic from a convex bicombing necessarily shares. Let X be a metric space. We call a curve γ : [a, b] → X straight (or a straight segment) if for every z ∈ X the function dz ◦ γ is convex, where dz = d(z, ·). In particular, for fixed s,t ∈ [a, b], taking z := γ(s) one gets the inequality

d(γ(s), γ((1 − λ)s + λt)) ≤ λ d(γ(s), γ(t)) for all λ ∈ [0, 1], whereas for z := γ(t) one obtains

d(γ((1 − λ)s + λt), γ(t)) ≤ (1 − λ) d(γ(s), γ(t)) for all λ ∈ [0, 1].

By taking the sum of these two inequalities one sees that straight curves are geodesics (of constant speed). The terminology is further justified by Theorem 3.3 below. In the proof of this result as well as in Example 3.5 we use Isbell’s injective hull construction [14], which we first recall (see [18] for a more detailed exposition). Given a metric space X, consider the vector space RX of arbitrary real valued functions on X and put

∆(X) := {f ∈ RX : f(x)+ f(y) ≥ d(x,y) for all x,y ∈ X}.

Call f ∈ ∆(X) extremal if there is no g ≤ f in ∆(X) distinct from f. The set E(X) of extremal functions is equivalently described as

RX E(X)= f ∈ : f(x) = supy∈X (d(x,y) − f(y)) for all x ∈ X , and it is easily seen that E(X) is a subset of

∆1(X) := {f ∈ ∆(X) : f is 1-Lipschitz}.

X Note that a function f ∈ R belongs to ∆1(X) if and only if

kf − dxk∞ = f(x) for all x ∈ X. (3.1) 8 D. Descombes & U. Lang

The metric (f, g) 7→ kf −gk∞ on ∆1(X) is thus finite, as kf −gk∞ ≤ kf −dxk∞ + kg − dxk∞ = f(x)+ g(x) for any x ∈ X. The set E(X) ⊂ ∆1(X) is equipped with the induced metric, and one has the canonical isometric embedding

e: X → E(X), e(x)= dx.

Isbell showed that (e, E(X)) is indeed an injective hull of X. That is, E(X) is an injective metric space, and every isometric embedding of X into another injective metric space factors through e. Returning to straight curves, we first observe that this property persists when we pass from X to E(X). 3.1 Lemma. Let γ : [a, b] → X be a straight curve in some metric space X. Regarding X as a subspace of its injective hull E(X) we then have that γ is also straight in E(X). Proof. By (3.1), the distance from an element f ∈ E(X) to a point x ∈ X ⊂ E(X) equals f(x). So we need to show that the function f ◦γ is convex. Given x := γ(s), y := γ(t), and w := γ((1 − λ)s + λt), where λ ∈ [0, 1], let ε> 0 and choose (using the extremality of f) a point v ∈ X such that f(v)+ f(w) ≤ d(v, w)+ ε. Since γ is straight in X, we have

d(v, w) ≤ (1 − λ) d(v,x)+ λ d(v,y).

Furthermore, d(v,x) ≤ f(v)+ f(x) and d(v,y) ≤ f(v)+ f(y). Combining these inequalities we get

f(v)+ f(w) ≤ f(v) + (1 − λ)f(x)+ λf(y)+ ε.

Since ε was arbitrary, this gives f(w) ≤ (1 − λ)f(x)+ λf(y), as desired.

Now let X = V be a normed real vector space. Isbell [15] and Rao [21] (see also [6]) showed that then the injective hull E(V ) has a structure with respect to which the isometric embedding e: V → E(V ) is linear. Since E(V ) is injective, collections of balls in E(V ) have the binary intersection property, so the Banach space E(V ) is also injective in the linear category by [19]. Then a theorem of Nachbin, Goodner, and Kelley [16] implies that E(V ) is isometrically isomorphic to the space C(M) of continuous functions, with the supremum norm, on some extremally disconnected compact Hausdorff space M. Summarizing, we may thus view V as a linear subspace of C(M), where M is such that E(V ) =∼ C(M). This fact will be used below. As usual, we call a curve γ : [a, b] → V in a vector space V linear if it is of the form t 7→ p + ty for some p,y ∈ V . This is obviously a local property. 3.2 Proposition. Let M be a compact Hausdorff space, and let γ : [a, b] → O be a curve in an open subset O of C(M). Then γ is straight (in O with the induced metric) if and only if it is linear. Convex geodesic bicombings and hyperbolicity 9

Proof. Clearly every linear curve is straight. For the other direction, we assume that γ : [0, 1] → O is a straight curve from 0 to y, where kyk∞ = l, and the closed 2l-neighborhood of γ([0, 1]) is contained in O. We have to show that the two functions γ(λ) and λy agree for every λ ∈ (0, 1). So fix λ ∈ (0, 1) as well as m ∈ M. For an arbitrary ε> 0, choose an open neighborhood U of m such that

|γ(λ)(m′) − γ(λ)(m)| < ε/2, |y(m′) − y(m)| < ε/2 for all m′ ∈ U. By Urysohn’s lemma there is a nonnegative function φ ∈ C(M) with φ|M\U ≡ 0 and kφk∞ = 2l. Put z± := γ(λ) ± φ and note that these are elements of O. Since γ is straight,

2l = kz± − γ(λ)k∞ ≤ (1 − λ)kz±k∞ + λkz± − yk∞. (3.2)

′ Now if f ∈ C(M) is a function satisfying kfk∞ ≤ l as well as f(m )

kf + φk∞ ≤ 2l + f(m)+ ε.

Taking f = γ(λ) and f = γ(λ) − y we get kz+k∞ ≤ 2l + γ(λ)(m) + ε and kz+ − yk∞ ≤ 2l + γ(λ)(m) − y(m)+ ε, respectively. (Note that kfk∞ ≤ l since γ is a geodesic from 0 to y.) Together with (3.2), this shows that

γ(λ)(m) − λy(m)+ ε ≥ 0.

Similarly, for f = −γ(λ) and f = y − γ(λ) we obtain kz−k∞ ≤ 2l − γ(λ)(m)+ ε and kz− − yk∞ ≤ 2l + y(m) − γ(λ)(m)+ ε, respectively, which gives

λy(m) − γ(λ)(m)+ ε ≥ 0.

Letting ε → 0 we conclude that γ(λ)(m) = λy(m), ending the proof as λ and m were arbitrary.

Combining these results we obtain the desired characterization of linear geodesics in normed spaces.

3.3 Theorem. A straight curve in an arbitrary normed space V is linear. Hence the bicombing of linear geodesics is the only convex bicombing on V .

Proof. As mentioned above, we can regard the normed space V as being a linear subspace of its injective hull E(V ), and we have E(V ) =∼ C(M) for some compact Hausdorff space M. Now straight curves in V are straight in C(M) by Lemma 3.1 and linear by Proposition 3.2.

One may ask whether this holds more generally for linearly convex subsets of normed spaces. The answer turns out to be negative. 10 D. Descombes & U. Lang

3.4 Example. Consider the Banach space ℓ∞([0, 1]) of bounded real valued func- tions on [0, 1], equipped with the supremum norm. The subset

C := {f ∈ ℓ∞([0, 1]) : f(0) + f(1) = 1, f is convex, and f ∈ ∆1([0, 1])} is compact and linearly convex. Hence there is the convex bicombing of linear geodesics in C. But there are more straight segments: the curve γ : [0, 1] → C, γ(t) = dt, is non-linear, and by (3.1) we have t 7→ kf − γ(t)k∞ = f(t) for every f ∈ C, which is a convex function by the definition of C.

We conclude this section with an example of a compact metric space that admits at least two distinct convex geodesic bicombings.

3.5 Example. Consider two geodesics α, β : [0, 1] → ℓ∞([0, 1]) from d0 to d1: the linear geodesic α(s) = (1−s)d0 +sd1, and the Kuratowski embedding β(t)= dt of [0, 1]. Let B ⊂ ℓ∞([0, 1]) be the bigon composed of these two geodesic segments. By (3.1) we have

kα(s) − β(t)k∞ = α(s)(t)= s + t − 2st. (3.3)

Since this last term is symmetric in s and t, there is an isometric involution ι: B → B that interchanges α and β. Furthermore, as ℓ∞([0, 1]) is injective, we can embed the injective hull E(B) of B into ℓ∞([0, 1]) (and identify it with its image), so that B ⊂ E(B) ⊂ ℓ∞([0, 1]). Now E(B) is not linearly convex in ℓ∞([0, 1]), but by retracting the linear geodesics with endpoints in E(B) to E(B) we obtain a conical geodesic bicombing σ on E(B) (compare Lemma 2.1). Note that since α was already linear, we have σd0d1 = α. Note further that E(B) is compact, because B is compact. Theorem 1.1 then yields a convex bicombingσ ¯ which, by Remark 2.4, still satisfiesσ ¯d0d1 = α. Finally, the involution ι extends uniquely to an isometry I of E(B) (see, for instance, Proposition 3.7 in [18]).

Mappingσ ¯ by I we get a convex bicombingτ ¯ on E(B) withτ ¯d0d1 = β, distinct fromσ ¯.

4 Combinatorial dimension

The example just described contrasts with Theorem 1.2, which we will prove in this section. First we discuss the structure of injective hulls of finite metric spaces and the notion of combinatorial dimension in more detail. Let X be a finite metric space, with |X| = n ≥ 1, say. The set ∆(X) ⊂ RX =∼ Rn is an unbounded polyhedral domain, determined by the finitely many linear inequalities f(x)+ f(y) ≥ d(x,y) for x,y ∈ X (in particular f ≥ 0). As X is finite, a function f ∈ ∆(X) is extremal if and only if for every x ∈ X there exists a point y ∈ X such that f(x)+ f(y) = d(x,y). So the injective hull is a polyhedral subcomplex of ∂∆(X) of dimension at most n/2. (It is not difficult to see that E(X) consists precisely of the bounded faces of ∂∆(X).) For n ≤ 5, the Convex geodesic bicombings and hyperbolicity 11 various possible combinatorial types of E(X) are depicted in Sect. 1 of [7] (where ∆(X) and E(X) are denoted PX and TX , respectively). To describe the structure of E(X) further, one may assign to every f ∈ E(X) the undirected graph with vertex set X and edge set

A(f)= {x,y} : x,y ∈ X, f(x)+ f(y)= d(x,y) .  Note that this graph has no isolated vertices (because f is extremal), but may be disconnected, and there is a loop {x,x} if and only if f(x) = 0, which occurs if and only if f = dx (by (3.1)). Call a set A of unordered pairs of (possibly equal) points in X admissible if there exists an f ∈ E(X) with A(f)= A, and denote by A (X) the collection of admissible sets. The family of polyhedral faces of E(X) is then given by {P (A)}A∈A (X), where

P (A)= {f ∈ ∆(X) : A ⊂ A(f)}, and where P (A′) is a face of P (A) if and only if A ⊂ A′. We define the rank rk(A) of A as the dimension of P (A). This number can be read off as follows. If f, g are two elements of P (A), then f(x)+ f(y) = d(x,y) = g(x)+ g(y) for {x,y} ∈ A, so f(y) − g(y) = −(f(x) − g(x)). Thus the difference f − g has alternating sign along all edge paths in the graph (X, A). It follows that there is either no or exactly one degree of freedom for the values of f ∈ P (A) on every connected component of (X, A), depending on whether or not the component contains a cycle of odd length. We call such components (viewed as subsets of X) odd or even A-components, respectively. The rank rk(A) = dim(P (A)) is then precisely the number of even A-components of X. (Here we have adopted the notation from [18], whereas [7] uses Kf = {(x,y) ∈ X × X : f(x)+ f(y)= d(x,y)} in place of A(f).) Now let again X be a general metric space. We recall that the combinatorial dimension of X, introduced by Dress, is the possibly infinite number

dimcomb(X) = sup{dim(E(Y )) : Y ⊂ X, |Y | < ∞}, see Theorem 9′ on p. 380 in [7]. This theorem contains in particular a character- ization of the inequality dimcomb(X) ≤ n in terms of a 2(n + 1)-point inequality, which may be rephrased as follows.

4.1 Theorem (Dress). Let X be a metric space, and let n ≥ 1 be an integer. The inequality dimcomb(X) ≤ n holds if and only if for every set Z ⊂ X with |Z| = 2(n + 1) and every fixed point free involution i: Z → Z there exists a fixed point free bijection j : Z → Z distinct from i such that

d(z, i(z)) ≤ d(z, j(z)). (4.1) z Z z Z X∈ X∈ 12 D. Descombes & U. Lang

The case n = 1 corresponds to the much simpler fact that dimcomb(X) ≤ 1 if and only if X is 0-hyperbolic in the sense of Gromov [11] or tree-like in the terminology of [7], that is, for every quadruple of points x,x′,y,y′ ∈ X,

d(x,x′)+ d(y,y′) ≤ max{d(x,y)+ d(x′,y′), d(x,y′)+ d(x′,y)}.

Theorem 4.1 follows from more general considerations in Sect. (5.3) of [7]. For convenience, we provide a streamlined and somewhat simplified argument in an appendix. However, this result will not be used in the present paper. By the results of [18], every proper metric space with integer valued metric that is discretely geodesic and δ-hyperbolic has finite combinatorial dimension. By contrast, the unit circle S in the Euclidean plane with either the induced (chordal) or the induced inner metric satisfies dimcomb(S) = ∞, as is seen by looking at the constant extremal function f = diam(S)/2, restricted to the vertices of a regular 2n-gon. Similarly, the metric bigon B constructed in Example 3.5 has dimcomb(B) = ∞: consider the function f defined by f(α(s)) = f(β(1 − s)) = kα(s) − β(1 − s)k∞/2 for s ∈ [0, 1]. Among the finite dimensional normed spaces, only those with a polyhedral norm have finite combinatorial dimension, equal to the number of pairs of opposite facets of the unit ball. The following proposition is the key observation for the proof of Theorem 1.2. We denote by B(x, r) the closed ball of radius r at x.

4.2 Proposition. Let X be a metric space of finite combinatorial dimension. Then for every pair of points x0,y0 ∈ X there exists a δ > 0 such that

d(x0,y0)+ d(x,y) ≤ d(x,y0)+ d(x0,y) for all pairs of points x ∈ B(x0, δ) and y ∈ B(y0, δ).

Proof. When x0 = y0, the triangle inequality in X gives the result. Now assume that x0 6= y0. Denote by F the collection of all real valued functions with finite support spt(f) ⊂ X such that f ∈ E(spt(f)), x0,y0 ∈ spt(f), and {x0,y0}∈ A(f), that is, f(x0)+ f(y0)= d(x0,y0).

Since dimcomb(X) < ∞, there exist an integer n and a function f ∈ F such that rk(A(g)) ≤ n = rk(A(f)) for all g ∈ F . Since x0 6= y0, we have n ≥ 1, thus f > 0. By restricting f to a smaller set if necessary, we can arrange that A(f) is the collection of n disjoint pairs {x0,y0}, {x1,y1},..., {xn−1,yn−1}. There exists a δ > 0 such that for all v ∈ {x0,y0} and w ∈ {x1,y1,...,xn−1,yn−1} we have d(v, w) > δ, f(v) ≥ δ, and

f(v)+ f(w) ≥ d(v, w) + 2δ. (4.2)

Note that d(x0,y0) = f(x0)+ f(y0) ≥ 2δ. Let x ∈ B(x0, δ) and y ∈ B(y0, δ). If x = x0 or y = y0 or x = y, the desired inequality holds. So assume that Convex geodesic bicombings and hyperbolicity 13

x0 6= x 6= y 6= y0. Then x,y,x0,y0,...,xn−1,yn−1 are pairwise distinct. Put a := d(x,y0) − f(y0) and b := d(x0,y) − f(x0). We have

a ≥ d(x0,y0) − d(x,x0) − f(y0)= f(x0) − d(x,x0) ≥ f(x0) − δ, (4.3) b ≥ d(x0,y0) − d(y,y0) − f(x0)= f(y0) − d(y,y0) ≥ f(y0) − δ.

Since f(x0),f(y0) ≥ δ, this gives in particular a, b ≥ 0 and hence

a + f(x0) ≥ δ ≥ d(x,x0), (4.4) b + f(y0) ≥ δ ≥ d(y,y0).

Furthermore, for every w ∈ {x1,y1,...,xn−1,yn−1}, combining (4.3) and (4.2) we obtain

a + f(w) ≥ f(x0)+ f(w) − δ ≥ d(x0, w)+ δ ≥ d(x, w), (4.5) b + f(w) ≥ f(y0)+ f(w) − δ ≥ d(y0, w)+ δ ≥ d(y, w).

Now, in the case that a + b < d(x,y), we could define a function g with finite support by putting g(w) := f(w) for w ∈ {x0,y0,...,xn−1,yn−1} and by choosing g(x) > a and g(y) > b such that g(x)+ g(y)= d(x,y). In view of (4.4) and (4.5), this function would satisfy

A(g)= {x0,y0},..., {xn−1,yn−1}, {x,y} , so that g ∈ F and rk(A(g)) = n + 1, in contradiction to the maximality of n. We conclude that d(x,y) ≤ a + b = d(x,y0)+ d(x0,y) − d(x0,y0).

Resuming the discussion of straight curves, we can now prove the following.

4.3 Proposition. Suppose that X is a metric space of finite combinatorial di- mension, and α, β : [0, 1] → X are two straight curves. Then the function s 7→ d(α(s), β(s)) is convex on [0, 1]. In particular, every pair of points in X is joined by at most one straight segment, up to reparametrization.

Proof. For s,t ∈ [0, 1], put h(s,t) := d(α(s), β(t)). Fix s0 ∈ (0, 1). Then it follows from Proposition 4.2 that there exists an ε > 0 such that [s0 − ε, s0 + ε] ⊂ [0, 1] and, for all s,t ∈ [s0 − ε, s0 + ε],

h(s0,s0)+ h(s,t) ≤ h(s,s0)+ h(s0,t). (4.6)

Now suppose that s0 − ε ≤ s

h(s,s0) ≤ (1 − λ) h(s,s)+ λ h(s,t), (4.7) h(s0,t) ≤ (1 − λ) h(s,t)+ λ h(t,t). 14 D. Descombes & U. Lang

Combining (4.6) and (4.7) we conclude that

h(s0,s0) ≤ (1 − λ) h(s,s)+ λ h(t,t).

Note that this holds whenever s0 − ε ≤ s 0 depends on s0. Since s 7→ h(s,s) = d(α(s), β(s)) is continuous on [0, 1], it follows easily that this function is convex.

Theorem 1.2 is an immediate corollary of Proposition 4.3. Note that a convex (or conical) geodesic bicombing σ can be restricted to any ball B = B(z, r) because B is σ-convex: x,y ∈ B implies σxy([0, 1]) ⊂ B.

5 Boundary at infinity

We now consider a complete metric space X with a convex and consistent (equiv- alently, conical and consistent) geodesic bicombing. Note that completeness is no restriction, as any conical bicombing may be extended to the completion of the underlying space. We define the geometric boundary and the closure of X by means of geodesic rays that are consistent with the given bicombing, and we equip the closure with a simple explicit metric. (Some general references for the analogous constructions in the case of CAT(0) spaces or Gromov hyperbolic spaces are [2, 4, 5, 11].) Then we prove Theorem 1.4. By a ray in X we mean an isometric embedding of R+ := [0, ∞). Two rays ξ, η in X are asymptotic if the function t 7→ d(ξ(t), η(t)) is bounded or, equivalently, the Hausdorff distance between the images of ξ and η is finite. In the presence of a convex and consistent bicombing σ on X we call a ray ξ : R+ → X a σ-ray if

ξ((1 − λ)s + λt)= σxy(λ) whenever 0 ≤ s ≤ t, x := ξ(s), y := ξ(t), and λ ∈ [0, 1]. It follows that for any two σ-rays ξ, η the map t 7→ d(ξ(a+αt), η(b+βt)) is convex on R+ for all a, α, b, β ≥ 0. The geometric boundary ∂σX is the set of equivalence classes of mutually asymptotic σ-rays in X, and we write

Xσ := X ∪ ∂σX.

For a unified treatment of the two parts of Xσ it is convenient to associate with every pair (o, x) ∈ X × X the eventually constant curve ̺ox : R+ → X,

σox(t/d(o, x)) if 0 ≤ t < d(o, x), ̺ox(t) := (5.1) (x if t ≥ d(o, x).

As a preliminary remark we note that for any basepoint o ∈ X and r ∈ R+, the radial retraction φr : X → B(o, r) defined by φr(x) := ̺ox(r) satisfies 2r d(φ (x), φ (y)) ≤ d(x,y) (5.2) r r d(o, x) Convex geodesic bicombings and hyperbolicity 15 whenever d(o, x) ≥ d(o, y) and d(o, x) > r. To see this, let s := r/d(o, x) and ′ ′ y := σoy(s). We have φr(x) = σox(s), and since σ is conical, d(φr(x),y ) ≤ ′ ′ ′ s d(x,y). Furthermore, d(y , φr(y)) = d(o, φr(y))−d(o, y ) ≤ d(o, φr(x))−d(o, y ) ≤ ′ ′ ′ d(φr(x),y ) and so d(φr(x), φr(y)) ≤ d(φr(x),y )+ d(y , φr(y)) ≤ 2s d(x,y). In particular, φr is 2-Lipschitz. The constant 2 is optimal, as one can see by looking at 2 the space ℓ∞ with 0 as the basepoint, the map φ1, and the points (1, 1), (1+ε, 1−ε). In order to prove that for every pair (o, x¯) ∈ X × ∂σX there is a σ-ray issuing from o and representing the classx ¯, we shall need the following estimate (compare Lemma II.8.3 in [4] for the case of CAT(0) spaces).

5.1 Lemma. Let X be a metric space with a convex and consistent geodesic bicombing σ, and let o, p ∈ X. Then for any σ-ray ξ with ξ(0) = p we have

2t d(o, p) d(̺ (t), ̺ (t)) ≤ ox oy T − d(o, p) whenever T > 2d(o, p), x,y ∈ ξ([T, ∞)), and 0 ≤ t ≤ T − 2 d(o, p).

′ Proof. Assume d(o, x) ≤ d(o, y) and let s := d(p,x)/d(p,y) and y := σoy(s). Since ξ is a σ-ray, we have x = σpy(s). As σ is conical,

d(x,y′) ≤ (1 − s) d(p, o) ≤ d(o, p). (5.3)

Now d(o, x) ≥ d(p,x) − d(o, p) ≥ T − d(o, p) and so d(o, y′) ≥ d(o, x) − d(x,y′) ≥ T − 2 d(o, p). Hence, for 0 ≤ t ≤ T − 2 d(o, p), we have φt(x) = ̺ox(t) and ′ φt(y )= ̺oy(t), and (5.2) gives the result.

The next result now follows by a standard procedure.

5.2 Proposition. Let X be a complete metric space with a convex and consistent geodesic bicombing σ. Then for every pair (o, x¯) ∈ X × ∂σX there is a unique σ-ray ̺ox¯ with ̺ox¯(0) = o that represents the class x¯. Furthermore, if r ≥ 0 and x = ̺ox¯(r), then ̺xx¯(t)= ̺ox¯(r + t) for all t ∈ R+. Proof. Let ξ be a σ-ray in the classx ¯, and let p := ξ(0). For n = 1, 2,... , put ̺n := ̺oξ(n). It follows from the preceding lemma that for every fixed t ≥ 0 the sequence ̺n(t) is Cauchy. In the limit one obtains a σ-ray ̺ox¯ issuing from o. As in (5.3), we have d(ξ(t), ̺n[t d(o, ξ(n))/n]) ≤ d(o, p) for 0 ≤ t ≤ n, hence d(ξ(t), ̺ox¯(t)) ≤ d(o, p) for all t ≥ 0. In particular, ̺ox¯ is asymptotic to ξ. Finally, ′ ′ if ̺ is another σ-ray issuing from o and asymptotic to ξ, then t 7→ d(̺oξ(t), ̺ (t)) ′ is a non-negative, bounded, convex function on R+ that vanishes at 0, so ̺ = ̺oξ. From this uniqueness property, the last assertion of the proposition is clear.

A natural topology on Xσ = X ∪ ∂σX may be described in different ways. First we fix a basepoint o ∈ X and consider the set

Rσ,o := {̺ox¯ :x ¯ ∈ Xσ} = {̺ox : x ∈ X} ∪ {̺ox¯ :x ¯ ∈ ∂σX} 16 D. Descombes & U. Lang

of generalized σ-rays based at o, given by (5.1) and Proposition 5.2. We equip Rσ,o with the topology of uniform convergence on compact subsets of R+. Clearly Rσ,o is compact if and only if X is proper, as a consequence of the Arzel`a–Ascoli theorem. The topology of Rσ,o agrees, under canonical identification, with the cone topology on Xσ, a basis of which is given by the sets

Uo(¯x,t,ε) := {y¯ ∈ Xσ : d(̺ox¯(t), ̺oy¯(t)) < ε} (5.4)

forx ¯ ∈ Xσ and t,ε > 0. Note that since φr is 2-Lipschitz, we have

d(̺ox¯(r), ̺oy¯(r)) ≤ 2 d(̺ox¯(t), ̺oy¯(t)) for all r ∈ [0,t]. (5.5)

Note also that Uo(x,t,ε) is just the open ball U(x, ε) in case t ≥ d(o, x)+ ε. It follows readily from the following lemma that this topology on Xσ is independent of the choice of basepoint o.

5.3 Lemma. Given o ∈ X, x¯ ∈ ∂σX, ε,t > 0, and p ∈ X, there exists T > 0 such that Up(¯x, ε/4, T ) ⊂ Uo(¯x,ε,t).

Proof. It follows from Lemma 5.1 and the construction of the ray ̺ox¯ in Proposi- tion 5.2 that if T is chosen sufficiently large, depending on d(o, p) and ε, t, and if x := ̺px¯(T ), then d(̺ox(t), ̺ox¯(t)) ≤ ε/4.

Likewise, for any pointy ¯ ∈ Xσ, if y := ̺py¯(T ) and d(p,y) is large enough, then

d(̺oy(t), ̺oy¯(t)) ≤ ε/4.

Now ify ¯ ∈ Up(¯x, T, ε/4), that is, d(x,y) < ε/4, then

d(̺ox(t), ̺oy(t)) ≤ 2 d(x,y) < ε/2

by (5.2), provided d(o, x), d(o, y) > t. We conclude that d(̺ox¯(t), ̺oy¯(t)) < ε and thusy ¯ ∈ Uo(¯x,t,ε) for sufficiently large T .

Next we equip Xσ with the metric defined by ∞ −s Do(¯x, y¯) := d(̺ox¯(s), ̺oy¯(s)) e ds (5.6) Z0 (compare Sect. 8.3.B in [11]). We have Do(o, x¯) ≤ 1, with equality if and only if x¯ ∈ ∂σX. For d(̺ox¯(t), ̺oy¯(t)) = a, (5.5) yields ∞ a t ∞ e−s ds ≤ D (¯x, y¯) ≤ 2a e−s ds + 2s e−s ds, 2 o Zt Z0 Zt and it follows easily from these estimates that the metric (5.6) induces the cone topology. Observe also that if d(o, x) = R and y = φr(x) for some 0 ≤ r ≤ R, then R ∞ −s −s −r −R Do(x,y)= (s − r) e ds + (R − r) e ds = e − e . Zr ZR Convex geodesic bicombings and hyperbolicity 17

In particular, for any σ-ray ξ issuing from o, the curve λ 7→ ξ(− log(1 − λ)), λ ∈ [0, 1), is a unit speed geodesic with respect to Do. Accordingly, for λ ∈ [0, 1], we define the radial retraction ψλ : Xσ → BDo (o, λ) = {y¯ : Do(o, y¯) ≤ λ} such that ψ1 = id and ψλ(¯x) := ̺ox¯(t) for λ < 1 and t := − log(1 − λ). In this latter case, the generalized ray ̺ox with endpoint x := ψλ(¯x) agrees with φt ◦ ̺ox¯, and since φt is 2-Lipschitz we obtain

∞ −s Do(ψλ(¯x), ψλ(¯y)) = d(φt(̺ox¯(s)), φt(̺oy¯(s))) e ds ≤ 2 Do(¯x, y¯) (5.7) Z0 for allx, ¯ y¯ ∈ Xσ. Thus ψλ is 2-Lipschitz with respect to Do. Finally, we note that ifx ¯ ∈ Xσ and λ, µ ∈ [0, 1], then clearly

Do(ψλ(¯x), ψµ(¯x)) ≤ |λ − µ|, (5.8) with equality when Do(o, x¯) ≥ max{λ, µ}. Now we turn to the result stated in the introduction.

Proof of Theorem 1.4. The map (¯x, λ) 7→ ψ1−λ(¯x) is continuous on Xσ × [0, 1] by (5.7) and (5.8) and contracts Xσ to o, so Xσ is contractible. To show that Xσ is locally contractible, we prove that in fact each of the sets Uo(¯x,t,ε) forx ¯ ∈ Xσ and t,ε > 0 (see (5.4)) is contractible. The same map −t as above, but restricted to Uo(¯x,t,ε) × [0, e ], contracts Uo(¯x,t,ε) to the subset U(̺ox¯(t), ε) ∩ B(o, t), which as an intersection of balls is σ-convex and hence itself contractible. To prove that ∂σX is a Z-set in Xσ, let an open set U in Xσ be given, and assume that U 6= ∅, Xσ. We want to find a homotopy h: U × [0, 1] → U from the identity on U to a map into U \ ∂σX such that the restriction of h to (U \ ∂σX) × [0, 1] takes values in U \ ∂σX. To this end, we note that the function µ: U → R defined by 1 µ(¯x) := D (o, x¯) − inf{D (¯x, y¯) :y ¯ ∈ X \ U} o 2 o σ is Lipschitz continuous with respect to Do and satisfies µ(¯x) < Do(o, x¯) for all x¯ ∈ U because U is open. It is then easy to see that h(¯x, λ) := ψmax{1−λ,µ(¯x)}(¯x) serves the purpose. It remains to show that Xσ is an AR (absolute retract). In case X has finite topological dimension, we have

dim(Xσ) = dim(X) < ∞.

Indeed, for every ε ∈ (0, 1), any open covering of the Do-ball BDo (o, 1 − ε) with −1 mesh ≤ ε gives rise, via ψ1−ε, to an open covering of Xσ with mesh ≤ 3ε and the same multiplicity (see, for instance, [5] for the definitions). It is then a stan- dard result that contractible and locally contractible metrizable spaces of finite 18 D. Descombes & U. Lang

dimension are absolute retracts (see [8]). However, finite dimension is not needed. Every metric space X with a conical geodesic bicombing is strictly equiconnected, as defined in [12], so X is an AR by Theorem 4 in that paper (see also [9]). Now, by Corollary 6.6.7 in [22] (a result attributed to O. Hanner and S. Lefschetz), Xσ is an ANR (absolute neighborhood retract) as it contains X as a homotopy dense subset; alternatively, one can give a short direct argument along the lines of Lemma 1.4 in [23]. Thus Xσ is a contractible ANR or, equivalently, an AR (see Corollary 6.2.9 in [22]).

As a concluding remark we note that, presumably, the spaces ∂σX and Xσ depend on the choice of σ. However, if X is a complete (not necessarily proper) Gromov hyperbolic space with a convex and consistent bicombing σ, then Xσ is independent of σ and furthermore ∂σX agrees with the boundary ∂∞X defined in terms of sequences converging to infinity. Indeed, a modification of Lemma 5.1 and Proposition 5.2 first shows that for every ray ξ in X there is a σ-ray asymptotic to ξ and issuing from the same point. As in the proof of Lemma 5.3 one can then conclude that distinct bicombings yield homeomorphic boundaries.

Appendix: Proof of Theorem 4.1

We use the notation from Sect. 4. First we show the “if” direction. Let Y ⊂ X be a finite set. If |Y |≤ 2n+1, the dimension of E(Y ) is at most n. Now suppose that |Y |≥ 2n + 2, f ∈ E(Y ), and Z ⊂ Y is a set with |Z| = 2n + 2 and with a fixed point free involution i: Z → Z such that {z, i(z)} ∈ A(f) for every z ∈ Z. By assumption, there exists a fixed point free bijection j : Z → Z such that j 6= i and

d(z, i(z)) ≤ d(z, j(z)). z Z z Z X∈ X∈ Since f(z)+ f(i(z)) = d(z, i(z)) and d(z, j(z)) ≤ f(z)+ f(j(z)), this gives

f(z)+ f(i(z)) ≤ f(z)+ f(j(z)). z Z z Z X∈ X∈ However, since both i and j are bijections, these two sums agree, so each of the inequalities d(z, j(z)) ≤ f(z)+ f(j(z)) must in fact be an equality, that is, {z, j(z)} ∈ A(f). There is at least one z ∈ Z such that j(z) 6= i(z), so the graph

with vertex set Z and edge set z∈Z{{z, i(z)}, {z, j(z)}} has at most n connected components. As this holds for every Z and i as above, we conclude that also the S graph (Y, A(f)) has no more than n components. Since f ∈ E(Y ) was arbitrary, this shows that the dimension of E(Y ) is less than or equal to n. Now we prove the other implication. Suppose that Z ⊂ X is a set with |Z| = 2n + 2, and i: Z → Z is a fixed point free involution. Let Z2 denote the set of all subsets of cardinality two of Z. The involution i selects a subset Convex geodesic bicombings and hyperbolicity 19

Zi := {{z, i(z)} : z ∈ Z} of Z2 of n+1 disjoint pairs. For every function h: Z → R we consider the set W (h) of all w : Z2 → R such that w ≤ 0 on Zi, w ≥ 0 on Z2 \Zi, and w({z, z′})= h(z) ′ z ∈XZ\{z} for all z ∈ Z. First we observe that if z ∈ Z, and if {x,y}∈ Zi is chosen such that z 6∈ {x,y}, then the function wz,{x,y} := (δ{x,z} +δ{y,z} −δ{x,y})/2 on Z2 belongs to W (δz), and wi(z),{x,y} − δ{z,i(z)} is in W (−δz). It follows that W (h) is non-empty for every h: Z → R. Put µ(h) := sup{S(w) : w ∈ W (h)}, S(w) := w({x,y})d(x,y); {x,yX}∈Z2 note that µ(h) > −∞ but possibly µ(h) = ∞. For wz,{x,y} as above we have S(wz,{x,y}) ≥ 0 by the triangle inequality, thus µ(δz) ≥ 0 for all z ∈ Z. Further- more, since 0 ∈ W (0), also µ(0) ≥ 0. In fact, if w ∈ W (0), then λw ∈ W (0) for all λ ≥ 0, so we have either µ(0) = 0 or µ(0) = ∞. In case µ(0) = ∞, choose w ∈ W (0) with S(w) > 0 such that no w′ ∈ W (0) with S(w′) > 0 has strictly smaller support. It follows that every nonzero v ∈ W (0) with support spt(v) ⊂ spt(w) satisfies S(v) > 0, for if λ> 0 is the maximal number with the property that |λv| ≤ |w|, then v′ := w − λv belongs to W (0) and spt(v′) is a strict subset of spt(w), so S(v′) ≤ 0 and λS(v) = S(w) − S(v′) > 0. (In fact λv = w, because v′ 6= 0 would likewise imply S(v′) > 0.) Since for fixed z the sum of the weights w({z, z′}) is zero, it is not difficult to see that there exist pairwise distinct points z0, z1,...,zl, with l ≥ 1, such that w({zk, i(zk)}) < 0 < w({i(zk), zk+1}) for k = 0,...,l, where zl+1 := z0. Then the function l

v := −δ{zk,i(zk)} + δ{i(zk),zk+1} Xk=0 on Z2 belongs to W (0), and spt(v) ⊂ spt(w). Hence S(v) > 0. This means that l l d(zk, i(zk)) < d(i(zk), zk+1). Xk=0 Xk=0 The points i(z0),...,i(zl) are distinct, so there is a well-defined map j : Z → Z such that j(i(zk)) = zk+1 for k = 0,...,l and j(z)= i(z) otherwise. Note that j is fixed point free since {i(zk), zk+1}∈ Z2 and i(z) 6= z for all z ∈ Z. Furthermore, j is injective because i is injective, z0,...,zl are distinct, and j(z) = i(z) = zk would imply j(z) = j(i(zk)) = zk+1 6= zk. Now it is clear that (4.1) holds, even with strict inequality. Note that this part of the proof does not use the assumption dimcomb(X) ≤ n. It remains to consider the case µ(0) = 0. Let h, h′ ∈ RZ . For all w ∈ W (h) and w′ ∈ W (h′) we have w + w′ ∈ W (h + h′), therefore µ(h)+ µ(h′) ≤ µ(h + h′); 20 D. Descombes & U. Lang in particular µ(h) ≤ µ(0) − µ(−h)= −µ(−h) < ∞. Put

µ(h) − µ(−h) ν(h) := . 2 We have µ(h) − µ(−h′) ≥ µ(h + h′) and µ(h′) − µ(−h) ≥ µ(h + h′), so

ν(h)+ ν(h′) ≥ µ(h + h′).

We have already observed that µ(δz) ≥ 0 for all z ∈ Z, hence −µ(−δz) ≥ 0 and ν(δz) ≥ 0. If {x,y}∈ Z2 \ Zi, then δ{x,y} ∈ W (δx + δy), thus

ν(δx)+ ν(δy) ≥ µ(δx + δy) ≥ d(x,y).

Similarly, if {x,y}∈ Zi, then −δ{x,y} ∈ W (−δx − δy), hence

ν(−δx)+ ν(−δy) ≥ µ(−δx − δy) ≥−d(x,y) and so ν(δx)+ν(δy)= −ν(−δx)−ν(−δy) ≤ d(x,y). Now it is clear that there exists Z a function f ∈ R such that f(z) ≥ ν(δz) for all z ∈ Z and f(x)+ f(y)= d(x,y) for all {x,y} ∈ Zi. Hence, f ∈ ∆(Z) and Zi ⊂ A(f), so in fact f ∈ E(Z). Since dimcomb(X) ≤ n by assumption, we have rk(A(f)) ≤ n, and there is no loss of generality in assuming that there is no g ∈ E(Z) with Zi ⊂ A(g) and rk(A(g)) > rk(A(f)). Since Z has at most n even A(f)-components, some such ′ component Z contains more than one of the n + 1 edges in Zi. Note that i maps Z′ onto Z′. We claim that for every z′ ∈ Z′ there is a z′′ ∈ Z′ \ {z′, i(z′)} such that {z′, z′′} ∈ A(f). If, to the contrary, there were a z′ ∈ Z′ with no such z′′, we could choose g ∈ RZ such that f(z′) > g(z′) > d(z′, z) − f(z) and g(z) = f(z) for all z ∈ Z \ {z′, i(z′)}, and g(i(z′)) = d(z′, i(z′)) − g(z′). This function would satisfy rk(A(g)) > rk(A(f)) and Zi ⊂ A(g), in contradiction to the ′ assumption on f. It follows easily that there exist z0, z1,...,zl ∈ Z , with l ≥ 1, such that {z0, i(z0)},..., {zl, i(zl)} are pairwise disjoint and {i(zk), zk+1}∈ A(f) for k = 0,...,l, where zl+1 = z0. Now

l l l d(zk, i(zk)) = f(zk)+ f(i(zk)) = d(i(zk), zk+1). k k k X=0 X=0 X=0

Putting j(i(zk)) = zk+1 and j(zk+1) = i(zk) for k = 0,...,l and j(z) = i(z) otherwise, we get a fixed point free involution j : Z → Z distinct from i such that (4.1) holds with equality.

Acknowledgments We thank Tadeusz Januszkiewicz, Alexander Lytchak, and Viktor Schroeder for helpful discussions. We acknowledge support from the Swiss National Science Foundation. Convex geodesic bicombings and hyperbolicity 21

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D. Descombes ([email protected]), U. Lang ([email protected]), Department of Mathematics, ETH Zurich, 8092 Zurich, Switzerland